According to the article "Nuclear shapes" by Renee Lucas the nucleus's shape is mainly modified by vibrational and rotational features happening within the cell. According to the article if i read correctly "near closed shells spherical shapes prevail, while between closed shells the large number of valence nucleons in orbit with large particle angular momentum leads to nuclei with large deformations leading them to not only maintain its shape but also alloying it to work.
Answer:
391.5 J
Explanation:
The amount of work done can be calculated using the formula:
- W = F║d
- where the force is parallel to the displacement
Looking at the formula, we can see that the mass of the object does not affect the work done on it.
Substitute the force applied and the displacement of the object into the equation.
- W = (87 N)(4.5 m)
- W = 391.5 J
The amount of work done on the object is 391.5 J in order to move it 4.5 meters with an applied force of 87 Newtons.
Answer:
108 km
Explanation:
The conversion factor between meters and feet is
1 m = 3.28 ft
So the second altitude, written in feet, can be rewritten in meters as
or in kilometers,
the first altitude in kilometers is
so the difference between the two altitudes is
Answer:
5.56 A
Explanation:
From the question,
Q = it.............. Equation 1
Where Q = charges, i = current, t = time.
Make i the subject of the equation
i = Q/t.............. Equation 2
Given: Q = 200 coulombs, t = 0.6 minutes = (0.6×60) seconds
Substitite these values into equation 2
i = 200/(0.6×60)
i = 5.56 A
Hence the magnitude of the current flowing through the circuit is 5.56 A
